Surface integral of vector field

[sandylam966]

Homework Statement

find ∫E.dS, where E = (Ar^2, Br (sinθ),C cosρ), over the outside conical surface S, given by 1≤r≤2, θ=\pi/3 (this is an open surface, excluding the end faces).

Homework Equations

The Attempt at a Solution

from the context I believe ρ is the plane polar angle on the x-y plane, so the surface is a slanted ring with width 1 around the z axis.
expressing S in vector v = ((√3 r cosρ)/2, (√3 r sinρ)/2, r/2), so ∫E.dS = ∫^{2pi}_{0}∫^{2}_{1} (Ar^2, (√3 Br)/2 ,C cosρ).((-√3 r cosρ)/4, (-√3 r sinρ)/4, 3r/4) drdρ, where I have taken the normal to the surface from outside to inside of the conical. Then, after the dot product each term in the integrand has a sine or cosine term, hence integrating ρ from 0 to 2pi will give zero.

the given answer is (7B pi)/2. where did I go wrong?

 

[bloby]

From the problem statement I would rather say (r, $\theta$, $\rho$) are spherical coordinates.

 

[sandylam966]

From the problem statement I would rather say (r, $\theta$, $\rho$) are spherical coordinates.

yes that's what I've used. using let θ=pi/3 I parameterise the surface in terms of ρ and r. did I make any mistake here?

 

[bloby]

The domain is right: 1<r<2 and 0<$\rho$<2 pi, but the vectors are expressed in the basis $\{e_{r},e_{\theta},e_{\rho}\}$.
The unit normal outside the surface is then very simple.
Can you find dS in this basis?

 

[sandylam966]

Yes, the vectors are expressed in the basis $\{e_{r},e_{\theta},e_{\rho}\}$.
The normal outside the surface is then very simple.

hmm I don't think I get it. partial differentiating v wrt r and ρ, then take the cross product of the 2, I got ((-√3 r cosρ)/4, (-√3 r sinρ)/4, 3r/4) , pointing into the ring. is this not the correct normal?

 

[bloby]

Ok, yes it is possible but way too complicate here. (E is expressed in the basis I mentioned above. You are expressing things in $\{e_x,e_y,e_z\}$ here so to take the dot product you would have to express E in $\{e_x,e_y,e_z\}$)
Keep things in shperical coordinates, it is much simpler here.
Remember
$e_{r}$: unite vector in the direction of growing r
$e_{\theta}$: unite vector in the direction of growing $\theta$
$e_{\rho}$: unite vector in the direction of growing $\rho$